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Can you help me with NET DISPLACEMENT

  1. Sep 3, 2008 #1
    Im having a bit of trouble understanding this and hoping someone can give me an idea how to understand and calculate net displacement of 3 different directions and distances?
     
  2. jcsd
  3. Sep 3, 2008 #2
    I'm pretty out of it (tired), but I'll see what I can explain. From my understanding, this is dealing with vectors. Now it's hard to tell without a diagram or more knowledge of the question but for each direction, break it down into vertical and horizontal directions, then add up all the vertical components and horizontal ones as well.
     
  4. Sep 3, 2008 #3
    ok lets use these values....130 due north, 50m due east, and 40 due south....??
     
  5. Sep 3, 2008 #4
    Alright, well the vertical (positive or negative) would be the north and south ones. So lets say up (north) is positive. How far from the start position are you if you go 40 units south, then go 130 units north? (Ignore the east direction for now)
     
  6. Sep 3, 2008 #5
    well if you go 40 south and then 130 north you would be 90 from the start position
     
  7. Sep 3, 2008 #6
    Im assuminh your going to tell me on so now we have 90m North and then 50 East. so then take 90^2 + 50^2 and you get your answer for the net displacement>?
     
  8. Sep 3, 2008 #7
    Basically, but don't forget the last stage of the theorem: the square root!
    So many marks are lost due to this!!
     
  9. Sep 3, 2008 #8
    Basically ya, this is what I understood from the question. Hopefully that's what you wanted to do. Then you may* need to find the angle of displacement.
     
    Last edited: Sep 3, 2008
  10. Sep 3, 2008 #9
    yea see i found the answer of 102 for the displacement but now i need to find the angle and thats where im stuck lol i dont know why i didnt say that in the first place...guess i just questioned my own answer but i have no clue how to find the angle?
     
  11. Sep 3, 2008 #10
    K, well i'll attempt this, draw the lines out on a paper. So draw one line from the start point (call this anything, I'll say C for explaining purposes) going to the right (east) 50 so lets say 5cm (even though in this case the drawing doesn't need to be exact). Now from there, draw a line going up (north) 90 (so 9cm).

    Now from C, draw a line to the tip of the 9cm line you drew. Lets call this line D, the vertical line A, and the horizontal line B.

    D is your displacement or resultant vector. To find the angle, you use trig. You can use either sin, cos, or tan*. I would use tan because you would be working with nicer numbers (90 and 50, not 102.xxxxxxx)

    So let me know if you need more help. Long description here, sorry.
     
  12. Sep 3, 2008 #11
    yea i did that but my question is and maybe you can help me understand a little better. I know Tan is Opp/Adj but how do i determine which is the opposite and the adjacent? Do i consider Opposite to be the line away from angle im trying to find?
     
  13. Sep 3, 2008 #12
    Should be actually inverse tan I believe, along with the options of inverse cos, and inverse sin

    tan x = opposite/adjacent
    x = inv.tan (opp/adj)

    or...

    x = inv.cos (adj/hyp)

    or...

    x = inv.sin (opp/hyp)
     
  14. Sep 3, 2008 #13
    hey awesome man thanks i greatly appreciate this! mind if i ask your help for one more?
     
  15. Sep 3, 2008 #14
    Ah k, so you have 2 angles, one up where line D meets line A (vertical line) and we'll call this angle phi. You also have an angle where line D meets line B (horizontal line and we'll call this angle theta. Then a 90 degree angle where A meets B. Now you can use either angle phi or theta.

    Lets use angle theta (line D meets line B) so the opposite would be the vertical line, and the adjacent line would be the horizontal line.

    If you were to use angle phi, you would have the 'opposite' being the horizontal, and the adjacent would be the vertical. The 'adjacent' means the line that is right next to the angle, so it helps "form" the angle.
     
  16. Sep 3, 2008 #15
    the sun is 30 degrees above the horizon. it makes a 52 m long
    shadow of a tall tree. how high is the tree?
     
  17. Sep 3, 2008 #16
    What did you get for the angle, I just want to see if you managed to get it. The angle where line D meets the horizontal? I'm hoping I did it right but you should get... 60.9 degrees... I think haha.

    You have to specify sometimes too on your diagram, or if you don't draw one with words. Now since I don't have one on here to show, I would say, the displacement vector is 102.9units and is 60.9 degrees from the horizontal.

    At least that is what I was told by past teachers.
     
  18. Sep 3, 2008 #17
    Best way I find to do most questions like this is to draw pictures. So, this will end up being another triangle.

    So start with a tree, so a line going up, now then I drew the "shadow", so another line at the base of the tree going to the right.

    Now draw a line connecting the lines to complete this triangle, and call this line something, I called it R.


    Now that you have the picture, sort of, you can extent the line R to go up and left of the 'tip of the tree' to imagine the sun off and up the the left. Hopefully you can picture this, now where should the 52m and the angle be placed?
     
  19. Sep 3, 2008 #18
    yes i just rounded up and my answer was 102m with a degree of 61 north of east
     
  20. Sep 3, 2008 #19
    Ok well since the shadow is base so that would be the 52m. and if the sun is to the left of the tree then angle directly above the 90 should be the 30 degree angle?
     
  21. Sep 3, 2008 #20
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